Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] s = br.readLine().split(" ");
		int n = Integer.parseInt(s[0]);
		s = br.readLine().split(" ");
		double[] probs = new double[n + 1];
		double[][] dp = new double[n + 1][n + 1];
		for (int i = 1; i <= n; i++)
			probs[i] = Double.parseDouble(s[i - 1]);

		dp[1][0] = 1 - probs[1];
		dp[1][1] = probs[1];

		for (int i = 2; i <= n; i++)
			dp[i][0] = (1 - probs[i]) * dp[i - 1][0];

		for (int i = 2; i <= n; i++) {
			for (int j = 1; j <= i; j++) {
				dp[i][j] = (dp[i - 1][j - 1] * probs[i]) + (dp[i - 1][j] * (1 - probs[i]));

			}
		}

//		for (int j = 0; j <= n; j++) {
//			for (int i = 1; i <= n; i++)
//				System.out.printf("%.4f %-10s", dp[i][j], "");
//			System.out.println();
//		}

		double sum = 0;
		for (int i = n; i >= (n + 1) / 2; i--)
			sum += dp[n][i];

		System.out.println(sum);
	}

}